A force `vec(F)=3hat(i)+4 hat(j)-3hat(k)` is applied at the point P, whose position vector is `vec(r) = hat(2i)-2hat(j)-3hat(k)`. What is the magnitude of the moment of the force about the origin ?

Options

A. 23 units
B. 19 units
C. 18 units
D. 21 units

Explanation

To find the magnitude of the moment of the force about the origin, we will follow these steps:

### Step 1: Identify the vectors
We are given:
- Force vector: \(\vec{F} = 3\hat{i} + 4\hat{j} - 3\hat{k}\)
- Position vector: \(\vec{r} = 2\hat{i} - 2\hat{j} - 3\hat{k}\)

### Step 2: Use the formula for the moment of force
The moment of the force \(\vec{M}\) about the origin is given by the cross product:
\[ \vec{M} = \vec{r} \times \vec{F} \]

### Step 3: Calculate the cross product
To calculate the cross product \(\vec{r} \times \vec{F}\), we can set up a determinant:
\[ \vec{M} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -2 & -3 \\ 3 & 4 & -3 \end{vmatrix} \]

Calculating this determinant:
\[ \vec{M} = \hat{i} \begin{vmatrix} -2 & -3 \\ 4 & -3 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & -3 \\ 3 & -3 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & -2 \\ 3 & 4 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. For \(\hat{i}\):
\[ \begin{vmatrix} -2 & -3 \\ 4 & -3 \end{vmatrix} = (-2)(-3) - (-3)(4) = 6 + 12 = 18 \]

2. For \(\hat{j}\):
\[ \begin{vmatrix} 2 & -3 \\ 3 & -3 \end{vmatrix} = (2)(-3) - (-3)(3) = -6 + 9 = 3 \]

3. For \(\hat{k}\):
\[ \begin{vmatrix} 2 & -2 \\ 3 & 4 \end{vmatrix} = (2)(4) - (-2)(3) = 8 + 6 = 14 \]

Putting it all together:
\[ \vec{M} = 18\hat{i} - 3\hat{j} + 14\hat{k} \]

### Step 4: Calculate the magnitude of the moment
The magnitude of \(\vec{M}\) is given by:
\[ |\vec{M}| = \sqrt{(18)^2 + (-3)^2 + (14)^2} \]
Calculating this:
\[ |\vec{M}| = \sqrt{324 + 9 + 196} = \sqrt{529} = 23 \]

### Final Answer
The magnitude of the moment of the force about the origin is \(23\) units.

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